Thread: Printf arguments

Hi,

I just started learning c/c++, and am asking this question out of pure curiosity.

Look at this code snippet:

#include <stdio.h>

int main()
{
int i=10;
short j=10;

unsigned int ui=10;
unsigned short uj=10;

printf("%i\n",i);
printf("%i\n",j);
printf("%d\n",i);
printf("%d\n",j);

printf("%i\n",ui);
printf("%i\n",uj);
printf("%d\n",ui);
printf("%d\n",uj);

return 0;
}

The output of this program is simply 8 "10"s

My question is this: What is the difference between %i and %d, and how does printf know in each case whether the variable being passed is signed or unsigned or a long or a short integer?

All I know is somehow it works, but I would be really interested to know where it gets this information from.

Thanks in advance.

Doubleanti is right in that %d and %i are the same for output (printf), but there are differences for input (scanf).

However, your code doesn't prove much, since you print everything using %d and %i (ie always signed decimal numbers), so even if you stored a 'negative' value in the unsigned ints (expecting some large unsigned number), it would still come out negative (%d would assume it to be signed).

In addition, your printing of short values is broken (it should be "%hd" not "%d"), and to print unsigned values, you have "%u" and "%hu" respectively.

> how does printf know in each case whether the variable being passed is signed or unsigned or a long or a short integer?
It doesn't, it assumes you've coded the % conversions correctly.

Thanks, thats made things a bit clearer, like I said. I still have lots to learn.